Theorem undif5 4447

Description: An equality involving class union and class difference. (Contributed by Thierry Arnoux, 26-Jun-2024.)

Assertion

Ref	Expression
undif5	⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) ∖ 𝐵) = 𝐴)

Proof of Theorem undif5

Step	Hyp	Ref	Expression
1		difun2 4444	. 2 ⊢ ((𝐴 ∪ 𝐵) ∖ 𝐵) = (𝐴 ∖ 𝐵)
2		disjdif2 4443	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∖ 𝐵) = 𝐴)
3	1, 2	eqtrid 2812	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) ∖ 𝐵) = 𝐴)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∖ cdif 3903   ∪ cun 3904   ∩ cin 3905  ∅c0 4286

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-nul 4287

This theorem is used by:  cyclnumvtx  30217  fressupp  33106  nn0diffz0  33211  elrgspnlem4  33631  mplidomlem  33983  vieta  34036  sucdifsn2  39194